# Find the original three terms

$$x$$, $$y$$, and $$\frac3{2x}$$ are non-zero terms in an arithmetic progression. If the third term is increased by $$1$$, the three terms now form a geometric progression. Find the original three terms.

Here are my steps:

1) The numbers were set into a proportion to find the common ratio. $$\dfrac xy = \dfrac{\tfrac3{2x}}y$$

2) The equation was set equal to $$0$$ such that $$0=\tfrac32x^2-y^2$$

I do not know how to move on from here or if the steps lead to the correct answer. How would you solve this problem?

• It's wrong. What you are doing is treating the original sequence as a geometric progression but what the question says is the original sequence is arithmetic. – cr001 Feb 11 at 11:46
• @cr001 In that case, do I find the common difference? – Bibliophile Feb 11 at 12:01
• Yes, that's correct. – cr001 Feb 11 at 12:03